Question: $f(x, y) = y^{2x}$ $\dfrac{\partial^2 f}{\partial y^2} = $
Taking a second order partial derivative is like taking a regular second order derivative. We take the partial derivative once, then we take another partial derivative. $\dfrac{\partial^2 f}{\partial y^2} = \dfrac{\partial}{\partial y} \left[ \dfrac{\partial f}{\partial y} \right]$ Let's differentiate! $\begin{aligned} \dfrac{\partial^2 f}{\partial y^2} &= \dfrac{\partial}{\partial y} \left[ \dfrac{\partial}{\partial y} \left[ y^{2x} \right] \right] \\ \\ &= \dfrac{\partial}{\partial y} \left[ 2x y^{2x-1} \right] \\ \\ &= 2x(2x - 1)y^{2x - 2} \end{aligned}$ Therefore, $\dfrac{\partial^2 f}{\partial y^2} = 2x(2x - 1)y^{2x - 2}$.